If \[a\] is a unit vector satisfying \[a\times r=b\], \[a\cdot r=c\] and \[a\cdot b=0\]. Then
determine the value of \[r\] .
(a) \[cb+\left( a\times b \right)\]
(b) \[ca+\left( a\times b \right)\]
(c) \[cb-\left( a\times b \right)\]
(d) \[ca-\left( a\times b \right)\]

Hint: In this question, we will first evaluate the value of \[a\times \left( a\times r \right)\] given that \[a\times r=b\]. Then using the triple cross product formula \[\overrightarrow{a}\times \left(
\overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}\cdot
\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}\cdot \overrightarrow{b} \right)\overrightarrow{c}\] for vectors \[\overrightarrow{a}\], \[\overrightarrow{b}\] and
\[\overrightarrow{c}\], we will then find the value of \[a\times \left( a\times r \right)\]. Then using the fact that \[a\] is a unit vector we have \[{{a}^{2}}=1\]. We will then substitute the value \[{{a}^{2}}=1\] and \[a\cdot r=c\] in the expression for \[a\times \left( a\times r \right)\] to get the value of \[r\].

Complete step-by-step answer:
We are given a unit vector \[a\].
\[\Rightarrow {{a}^{2}}=1\]
Since we have \[a\times r=b\], thus on evaluating the value of \[a\times \left( a\times r \right)\] by
substituting \[a\times r=b\] we get
\[a\times \left( a\times r \right)=a\times b\]
Now we know that for vectors \[\overrightarrow{a}\], \[\overrightarrow{b}\] and
\[\overrightarrow{c}\], then the triple cross product of vectors \[\overrightarrow{a}\],
\[\overrightarrow{b}\] and \[\overrightarrow{c}\]is given by
\[\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left(
\overrightarrow{a}\cdot \overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}\cdot
\overrightarrow{b} \right)\overrightarrow{c}\]
Using the above formula for the triple cross product in \[a\times \left( a\times r \right)\], we get that
\[a\times \left( a\times r \right)=\left( a\cdot r \right)a-\left( a\cdot a \right)r\]
Now, using \[a\cdot r=c\] in the above equation we have
\[\begin{align}
& a\times \left( a\times r \right)=\left( a\cdot r \right)a-\left( a\cdot a \right)r \\
& =ca-{{a}^{2}}r
\end{align}\]
Since \[a\times \left( a\times r \right)=a\times b\] , thus we have
\[ca-{{a}^{2}}r=a\times b...........(1)\]
Also \[{{a}^{2}}=1\]where \[a\]is a unit vector.
Therefore substituting the value \[{{a}^{2}}=1\] in equation (1) , we get
\[ca-r=a\times b\]
We will now calculate the value of \[r\] by rearranging the terms of the above equation.
By taking vector \[r\] to the right and taking \[a\times b\] to the left side of the equation we get
\[ca-\left( a\times b \right)=r\]
Therefore the value of \[r\] is given by \[ca-\left( a\times b \right)\]

So, the correct answer is “Option (d)”.

Note: In this problem, we have also use the definition of cross product and dot product to get the
desired value of \[r\]. We have \[a\times b=ab\sin \theta \] where \[\theta \] is the angle between \[a\] and \[b\].
Also \[a\cdot b=ab\cos \theta \]. Then using the fact that \[a\cdot b=0\], we say that vectors \[a\] and \[b\] are orthogonal to each other. That is the vectors \[a\] and \[b\] are perpendicular to each other.